Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 3 / 5 / a

The orbits of a normal subgroup form a block system: $g(N\omega )=N(g\omega )$, so every group element permutes these orbits. By primitivity the orbits are either all singletons or one whole orbit. In the singleton case $N$ fixes every point, giving $N\le G_{[\Omega ]}$.

In the other case $N$ is transitive. For each $g\in G$ choose $n\in N$ with $n\alpha =g\alpha$. Then $n^{-1}g\in G_{\alpha }$, so $G=N{G}_{\alpha }$. Work in the quotient $G/N$. If $g=nh$ with $h\in G_{\alpha }$, the image of $A^g=g^{-1}Ag$ is the image of $A^h$, which equals the image $\overline{A}$ of $A$ because $A◃G_{\alpha }$. All conjugates of $A$ thus have the same image in $G/N$.

By hypothesis those conjugates generate $G$, so the quotient is generated by the single subgroup $\overline{A}$, and equals it. Since $A$ is abelian, $G/N$ is abelian. The kernel of a group homomorphism of a group homomorphism to an abelian group contains the commutator subgroup, yieldingThis is the normal-subgroup argument behind the Iwasawa simplicity lemma. Faithfulness is not needed for this version: retaining the pointwise kernel of a group homomorphism as the second alternative is essential. In a faithful primitive group action it says every nontrivial normal subgroup contains $G^{\prime }$.